Results 1 to 8 of 8

Math Help - Quadratic Inequalities

  1. #1
    Junior Member
    Joined
    Feb 2010
    Posts
    25

    Quadratic Inequalities

    5a) Find the values of p for which the equation x^2 + px + 8 = p has one positive and one negative root [p > 8]

    I have found the determinant to be p^2 + 4p - 32
    Please advise how to continue - the answer is not obtained by taking determinant > 0

    b) Find the positive value of k for which the equation 6x^2 - 2kx + k = 0 has 2 real roots, one being thrice the other

    I have found the determinant to be k^2 - 6k
    Please advise how to proceed

    c) Given that A and B are the roots of the euqation x^2 = 3x + 5, find the value of 1/A^2 + 1/B^2 and show that A^4 = 57A + 70

    I have found the answer to be 19/25, but I've no idea how to show the equation
    Please help!

    Thanks a lot
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Member
    Joined
    Dec 2009
    Posts
    226
    For parts (a) and (b), the determinants must be greater than 0 for the quadratic equations in x to have two real solutions. Therefore, it's a matter of solving the inequalities p^2 + 4p - 32 > 0 and k^2 - 6k > 0.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member
    earboth's Avatar
    Joined
    Jan 2006
    From
    Germany
    Posts
    5,830
    Thanks
    123
    Quote Originally Posted by Drdj View Post
    5a) Find the values of p for which the equation x^2 + px + 8 = p has one positive and one negative root [p > 8]

    I have found the determinant to be p^2 + 4p - 32 <--- this term is called discriminant
    Please advise how to continue - the answer is not obtained by taking determinant > 0

    ...
    1. To solve the equation for x use the quadratic formula:

    x^2+px+8-p=0

    x = \dfrac{- p + \sqrt{p^2 +4p-32}}2~\vee~ x=\dfrac{- p - \sqrt{p^2 + 4p-32}}2

    2. Solve the inequalities for p:

    \dfrac{- p + \sqrt{p^2 +4p-32}}2 > 0 ~\wedge~ \dfrac{- p - \sqrt{p^2 +4p-32}}2 <0

     \sqrt{p^2 +4p-32} > p ~\wedge~  - \sqrt{p^2 +4p-32} < p

    p\le-8\vee p >8~~~~~~\wedge~~~~~~p\ge 4~\implies~p > 8

    3. Now solve the inequalities for p:

    \dfrac{- p + \sqrt{p^2 +4p-32}}2 < 0 ~\wedge~ \dfrac{- p - \sqrt{p^2 +4p-32}}2 > 0

    You should come out with p \le -8
    Last edited by earboth; April 9th 2011 at 12:52 PM.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member
    Joined
    Mar 2010
    Posts
    715
    Thanks
    2
    Quote Originally Posted by Drdj View Post
    c) Given that A and B are the roots of the euqation x^2 = 3x + 5, find the value of 1/A^2 + 1/B^2 and show that A^4 = 57A + 70
    x^2 = 3x+5 \iff x^2-3x-5 = 0.
    By Vieta's, a+b = 3 and ab = -5, and

    \displaystyle \frac{1}{a^2}+\frac{1}{b^2} = \frac{b^2+a^2}{a^2b^2} = \frac{(a+b)^2-2ab}{(ab)^2}.
    Last edited by TheCoffeeMachine; April 9th 2011 at 01:01 PM.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Junior Member
    Joined
    Feb 2010
    Posts
    25
    Thanks 'TheCoffeeMachine'
    Can you help me with the 2nd part - show that A^4 = 57A + 70 ?
    Thanks
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Junior Member
    Joined
    Feb 2010
    Posts
    25
    Thanks for your help, earboth, though I don't quite understand your method
    Is there an alternative way, or could you elaborate?
    Apologies
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Junior Member
    Joined
    Feb 2010
    Posts
    25
    Thanks for your help, earboth, though I have some queries about your method

    1.How do you know -p/2 - sqrt(p^2 + 4p - 32)/2 is <0? Why not >0?
    2.How do you solve sqrt(p^2 + 4p - 32) > p?
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Super Member
    Joined
    Mar 2010
    Posts
    715
    Thanks
    2
    Quote Originally Posted by Drdj View Post
    Thanks 'TheCoffeeMachine'
    Can you help me with the 2nd part - show that A^4 = 57A + 70 ?
    Thanks
    I'm sorry, I'm not sure how the second part follows. Perhaps they want you to find a and b explicitly first?
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Quadratic Inequalities
    Posted in the Algebra Forum
    Replies: 1
    Last Post: April 9th 2011, 10:23 AM
  2. Quadratic Inequalities
    Posted in the Algebra Forum
    Replies: 1
    Last Post: April 2nd 2011, 12:05 AM
  3. Quadratic Inequalities.
    Posted in the Algebra Forum
    Replies: 4
    Last Post: April 26th 2010, 06:14 PM
  4. Quadratic inequalities
    Posted in the Algebra Forum
    Replies: 5
    Last Post: December 13th 2008, 08:07 PM
  5. Replies: 1
    Last Post: June 12th 2008, 10:30 PM

Search Tags


/mathhelpforum @mathhelpforum